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Understanding \(f(z,\theta) = e^{e^{i\theta}z} - 1\) - JmsNxn - 12/10/2022
Hey, everyone.
So I've recently started to pay sole attention to:
\[
f(z) = e^z-1\\
\]
And the different petals this object represents (canonically there are 2 petals here). From this identification, I can create as good as possible iterations. Similarly, I managed to produce the same asymptotic series Gottfried wrote of \(f^{\circ 0.5}(z)\) about \(z=0\). My construction, although much slower and less accurate and precise compared to Gottfried, was much simpler. The trouble I want to talk about now is a tad different.
Let's write:
\[
f_\theta(z) = e^{e^{i\theta}z}-1\\
\]
For example, if \(\theta = \pi\):
\[
f_\pi(z) = e^{-z} - 1\\
\]
What is the fractional iteration of this beast!?
Well, if we reduce using basic mathematics this can be found. Let us say that:
\[
f^{\circ 0.5}(f_{\pi}(z)) = f^{\circ 1.5}(-z)\\
\]
Which is absolutely true. Further more, one can find that:
\[
f_\pi(f^{\circ 0.5}(z)) = f^{1.5}(-z)\\
\]
How does this relate to the iteration of \(f_\theta\) for \(\theta \in \mathbb{R}\). How does this relate across neutral fixed points...
Think of this, as us applying \(e^{i\theta}\) while we apply \(f\), and doing so synchronously. We are taking \(f_0 : z \to e^z -1\) and \(\theta: z \to e^{i\theta}z\). Somehow we are taking this in a manner to produce \(f^{\circ 0.5}_\theta\). We are taking the half root of both functions, while doing neither.
To begin this foray, we can say that \(f_\pi(f_\pi(z)) = z + O(z^2)\). Which means, if we iterate \(f_\pi\), we can get a nice function, which has \(1\) as it's multiplier. The trouble here being, any iterate of this, will be an iterate of \(f^{\circ 2}_\pi\) and not an iterate of \(f_\pi\).
So, how do we write an iterate of \(f_\theta(z)\)?
I hope to answer this shortly, but I'm not perfect here. This result is only true for \(|z| < \delta\), while additionally, is not holomorphic at \(z=0\), and additionally has branch cuts stemming from this point.
I am currently making graphs, and checking my code. but I believe I can explain all of this using the mellin transform and the appropriate iterations.
One of the key observations, which is entirely non obvious, is that:
\[
\vartheta_\pi(x,z) = \sum_{n=0}^\infty f_{\pi}^{\circ n}(z)\frac{(-x)^n}{n!}\\
\]
Does not have decay as \(x \to \infty\)--instead it has this decay as \(x \to -\infty\). The affectation of adding a \((-1)\) in the function argument, causes the taylor series to rotate its domain of convergence under the mellin transform. I hope to explain this better.
Write:
\[
\vartheta_\theta(x,z) = \sum_{n=0}^\infty f_{\theta}^{\circ n}(z)\frac{(-x)^n}{n!}\\
\]
This function has decay for \(\arg(x) = \theta\). And in neighborhoods of this. Whereby, results in mellin, describe results in Abel.
I apologize, I hope to make all of this work better. I just do not have the time as of now.
A helpful way to visualize this is with \(f = e^z-1\). To begin:
\[
f^{\circ n}(z) = f^n\\
\]
Are purely positive, and monotone. If I write:
\[
f_\pi^{\circ n} = f^n_\pi\\
\]
Then this monotone nature no longer appears.
When I write \(e^{-1} -1 < 0\), this appears throughout. When I write:
\[
e^1 -1 > 1
\]
This appears throughout. When I adapt this to taylor series we can write:
\[
\sum_{n=0}^\infty f^{\circ n} \frac{(-x)^n}{n!}\\
\]
Where the coefficients are monotone, and therefore the oscillation is only formed by \((-x)^n\). This allows us to take a mellin transform, because everything is nice. When I do this with \(f_\pi\), things are different:
\[
\vartheta_\pi(x) = \sum_{n=0}^\infty f_\pi^{\circ n} \frac{(-x)^n}{n!}\\
\]
Using this function, this does not have a nice decay as we increase \(x \to \infty\). Which is because \(f_\pi^{\circ n}\) looks like \((-1)^n\) itself. It oscilates in a similar manner. Where by, we must take \(x \to - \infty\), and then the integral converges.
FFS I'm tired boys...
To be straight with you guys, we have to think like Ramanujan:
\[
f^{\circ n}(z) = 1^n
\]
whereby:
\[
\vartheta(x) = e^{-x}\\
\]
From which:
\[
f_\pi^{\circ n}(z) = (-1)^n\\
\]
and:
\[
\vartheta_\pi(x) = e^x\\
\]
Where, at this point, all we have to do is change the direction of our integral... And this follows for fucking every direction... We can perform the iteration no matter what, but we have to write the integral correctly. First of all:
\[
\int_0^\infty \vartheta_0(x)\,dx\\
\]
converges, but then to do this when we change the multiplier:
\[
\int_{0}^{-\infty} \vartheta_\pi(x)\,dx\\
\]
We have to move the domain of integration, as we move our petals/initial points, ALL THIS FUCKING
BULLSHIT!!!!!! If I were to take \(\vartheta_{\pi/2}\) then I'd write something like:
\[
\int_0^{-i\infty}\vartheta_{\pi/2}(x)\,dx\\\
\]
And write various transforms attached to this (AGAIN, THIS MEANS WE BELONG TO A HILBERT SPACE). This is again no different than the identity:
\[
\int_0^{-i\infty} e^{-ix}\,dx\\
\]
And the fact this object converges. We typically attach this with a mellin transform, whereby:
\[
e^{-\pi i s /2} \Gamma(s) = \int_0^{-i\infty} e^{-ix}x^{s-1}\,dx\\
\]
We are doing something deeply similar; just without the obvious beauty of the Gamma function.
I really want to hear you guys out, but I'm just lost at how much you guys are forgetting. And ignoring the "Schrodinger" interpretation has left me blind sided. I plan to write a broken, but fairly well interpreted program. And all it uses is basic code.
EDIT:
A good way to conceptualize this, is if \(a_n \in \mathbb{R}^+\), and it has good enough decay, then:
\[
\vartheta(x) = \sum_{n=0}^\infty a_n \frac{(-x)^n}{n!}\\
\]
Can look a lot like \(e^{-x}\). When \(b_n = (-1)^n a_n\), then:
\[
\vartheta_\pi(x) = \sum_{n=0}^\infty b_n \frac{(-x)^n}{n!} \approx e^x\\
\]
BOTH OF THESE FUNCTIONS ARE MELLIN TRANSFORMABLE!!!!!
We just need to take the mellin transform in different manners.........
RE: Understanding \(f(z,\theta) = e^{e^{i\theta}z} - 1\) - JmsNxn - 12/12/2022
To begin, I'll present the first non obvious theorem, for this; I will start with \(f^{\circ 2}_\pi(z)\). So if \(f_\pi(z) = e^{-z}-1\), \(f^{\circ 2}_\pi = f_\pi(f_\pi(z))\) And by such a manner, I will deconstruct the function \(\vartheta_\pi\). As we begin we have:
\[
H_\pi(x,z) = \sum_{n=0}^\infty f^{\circ 2(n+1)}_\pi(z) \frac{(-x)^n}{n!}
\]
Now, as noted above, there's an abel function for \(f^{\circ 2}_\pi\), simply by nature of it having a neutral fixed point at \(z=0\), and has a multiplier \(\lambda = 1\). And therefore it has an iteration \((f^{\circ 2})^{\circ s}_\pi\). Which we can write as:
\[
\Gamma(1-s) (f^{\circ 2})^{\circ s}_\pi(z) = \int_0^\infty H_\pi(x,z)x^{s-1}\,dx
\]
We write \((f^{\circ 2})^{\circ s}_\pi\), because this powering operation is non-associative. The value \((f^{\circ 2})^{\circ 1/2}_\pi \neq f\). Instead, we solely have:
\[
\begin{align}
F(s,z) &=(f^{\circ 2})^{\circ s}_\pi(z)\\
F(1,z) &= f^{\circ 2}_\pi(z)\\
F(s+y,z) &= F(s,F(y,z))
\end{align}
\]
We now write, our first alteration, by which we wish to use this expansion to get the true \(f^{\circ s}_\pi\). We begin this by writing an integral transform:
\[
Q(x,z) = \frac{1}{2\pi i} \int_{-\frac{1}{2}-i\infty}^{-\frac{1}{2} + i \infty} \frac{\Gamma(1-s)\Gamma(s)}{\Gamma(1-2s)} (f^{\circ 2})^{\circ s}_\pi(z)x^{-2s}\,ds
\]
IF (Now this is a big if, which I plan on justifying, but momentarily, follow me--it is a slow convergence but converges upon massaging) this object converges it converges exactly to:
\[
Q(x,z) = \sum_{n=1}^\infty f^{\circ 2(n+1)}_\pi(z) \frac{(-1)^nx^{2n}}{(2n!)}
\]
But this is precisely:
\[
\frac{\vartheta_\pi(ix,z) + \vartheta_\pi(-ix,z)}{2} - f_\pi(z)= Q(x,z)
\]
Whereby this is mellin transformable--and puts us exactly at the point where the mellin transform of \(\vartheta_\pi\) can be found, up to a multiplicative factor. Explaining this is pretty advanced, so I'm going to start with an example every single one of you have seen--though perhaps not in this late staged format.
TRIGGER WARNING: This puts us right next door to Hard Analytic Number Theory, and additionally the work used to make the Riemann Hypothesis. I apologize, I was always one of those hubris young kids/nerds who studied everything in and around the Riemann hypothesis--by such, I've developed much of the same tools. I'm a nerd for this shit. We're like 1/2 half way before relating Hard Analytic Number Theory to complex dynamics.
----------------------------------------------------------------
Let's begin by writing:
\[
e^{-x} = \sum_{n=0}^\infty \frac{(-x)^n}{n!}
\]
This when applied under the mellin transform:
\[
\Gamma(1-s) = \int_0^\infty e^{-x}x^{-s}\,dx
\]
Now first of all, we know that:
\[
h(s) = 1
\]
Is a bounded function on \(\mathbb{C}_{\Re(s) > 0}\). (Like duh--but note in this proof we only use that it is bounded in this half plane, that's all we care about).
So we can instead write this as:
\[
\Gamma(1-s)h(s) = \int_0^\infty e^{-x}x^{-s}\,dx
\]
Now let's write:
\[
q(x) = \frac{1}{2\pi i} \int_{-\frac{1}{2}-i\infty}^{-\frac{1}{2}+i\infty} \frac{\Gamma(1-s)\Gamma(s)}{\Gamma(1-2s)}x^{-s}\,ds
\]
We can start, by proving this converges. It's not as hard as you might ask:
\[
\Gamma(x+iy) = O(e^{ -\frac{\pi}{2} |y|} |y|^{x-1/2})
\]
And this result holds asymptotically if the constant of proportion is chosen correctly. Whereby, we can also say:
\[
\frac{1}{\Gamma(x+iy)} = O(e^{ \frac{\pi}{2} |y|} |y|^{-x+1/2})
\]
Therefore:
\[
\frac{\Gamma(1-x-iy)\Gamma(x+iy)}{\Gamma(1-2x-2iy)} = O(\frac{|y|^{x-1/2}|y|^{1-x-1/2}}{|y|^{1-2x-1/2}}) = O(|y|^{2x - 1/2})
\]
We have set \(\Re(s) = x = -1/2\) in this integral, thereby.
\[
\frac{\Gamma(1-s)\Gamma(s)}{\Gamma(1-2s)} = O(|y|^{-\frac{3}{2}})
\]
.......................................
Now what does this object converge to?
\[
q(x) = \cos(\sqrt{x}) -1 = \sum_{n=1}^\infty \frac{(-1)^nx^n}{2n!}
\]
Where, as we noted above:
\[
q(x^2) = \frac{e^{ix} + e^{-ix}}{2} - 1 = \sum_{n=1}^\infty \frac{(-1)^n x^{2n}}{2n!}
\]
Now, when we take, for \(-1 <\Re(s) < 0\):
\[
\int_0^\infty q(x)x^{s-1}\,dx = \frac{\Gamma(1-s)\Gamma(s)}{\Gamma(1-2s)}
\]
Which through plain substitution:
\[
\int_0^\infty \left( \cos(u) - 1\right)u^{s-1}\,du = \cos(s\frac{\pi}{2})\Gamma(s)
\]
Where this is the duplication formula of the Gamma function in action--though with no obvious mention of anything of Gauss. We attribute this to Legendre, but to see the Cosine function, it's helpful to look at the general case by Gauss.
I am simply trying to do the same thing, but instead of the transformation with \(1 \to -1\), I am trying to write \(e^z -1 \to e^{-z}-1\). Surprisingly, under these mellin transforms, this is always possible. Though, the real trick becomes "How we take the mellin transform".
Anyway, that's all you'll have from me tonight. I'm going to use this thread as a note dump primarily; and I plan to expand on my previous rough paper. I don't have the time at the moment to develop these arguments fully, but numbers don't lie
![[Image: biggrin.gif]](https://math.eretrandre.org/tetrationforum/images/smilies/biggrin.gif){kind=small}
'I SCREWED UP SOME OF THE INDICES IN THIS POST
RE: Understanding \(f(z,\theta) = e^{e^{i\theta}z} - 1\) - JmsNxn - 12/12/2022
Also, to give an idea of what I am going to do:
\[
\frac{d^{s}}{dw^s}\Big{|}_{w=0} \frac{e^{iw} + e^{-iw}}{2} -1 = \frac{i^s + (-i)^s}{2}= \cos(\frac{\pi s}{2})\,\,\text{when}\,\,0 < \Re(s) < 1
\]
Who's to say the same thing doesn't happen with \(\vartheta_\pi\)
RE: Understanding \(f(z,\theta) = e^{e^{i\theta}z} - 1\) - JmsNxn - 12/22/2022
Let's get some notation clear right now.
We will call \(\mathcal{P}\) an attracting petal of \(f(z)\) at \(f(0) = 0\) where \(|f'(0)| = 1\). We will call \(\cup \mathcal{P}\) the collection of all attracting petals. We will say \(f\) is an entire function (mostly for convenience). From which, we have for all \(\mathcal{P} \subset \mathbb{C}\) that are petals:
\[
\lim_{n\to\infty} f^{\circ n} \Big{|}_{\mathcal{P}} \to 0\\
\]
Whereby, we are guaranteed now, that the union of such domains:
\[
\lim_{n\to\infty} f^{\circ n} \Big{|}_{\cup\mathcal{P}} \to 0\\
\]
Identify that \(f : \cup \mathcal{P} \to \cup \mathcal{P}\). Recall a "domain" is an open and connected set in \(\mathbb{C}\). \(\mathcal{P}\) is a domain, the set \(\cup \mathcal{P}\) is not a domain, it is a collection of domains... There's a bunch of branch cuts/julia set shit which separates everything.
As I'm too lazy to write the latex code every time:
\[
\frac{d^s}{dw^s}h(w) = H(s) = \frac{d^s}{dw^s}h(w)\Big{|}_{w=0}\\
\]
This is just a fancy mellin transform.
By the above renditions, we know that:
\[
\vartheta(w,z) = \sum_{n=0}^\infty f^{\circ n+1}(z) \frac{w^n}{n!}\\
\]
Is holomorphic for:
\[
\vartheta(w,z) : \mathbb{C} \times \cup\mathcal{P} \to \mathbb{C}\\
\]
Now, what we're trying to show; is that \(\frac{d^s}{dw^s} \vartheta(w,z)\) always exists.
This is highly non trivial. We know when \(f(z) = e^z-1\) and \(\cup \mathcal{P} = \mathbb{C}/[0,\infty)\); this absolutely happens. When we have a negative:
\[
f_-(z) = f(-z) = e^{-z}-1
\]
This instantly becomes a non obvious problem. The domain \(\cup \mathcal{P}\), now has a cycle permutation, for everytime we apply that negative. If we take:
\[
f_2(z) = e^{iz} - 1\\
\]
We now have a permutation of order 4. But no matter what these things shuffle around. They still produce holomorphic \(\vartheta\). And they produce integrals, though they may be on exotic domains.
To take \(f_-\) for example. The manner we take this integral is rather strange. We have our function:
\[
\vartheta(w,z): \mathbb{C} \times \cup\mathcal{P} \to \mathbb{C}\\
\]
And then we have the standard abel function on the attracting cup petal \(\cup\mathcal{P}\):
\[
\alpha(f_-(z)) = \alpha(z) + 1\\
\]
We can restrict this right away to a connected petal \(K \subset \cup \mathcal{P}\). The question, is whether the differintegral can do this.
To start this debauchery:
\[
b(w) = \sum_{k=0}^\infty f_-^{\circ 2(k+1)} \frac{w^k}{k!}\\
\]
Which:
\[
\frac{d^s}{dw^s} b(w) = \left(f_- \circ f_-\right)^{\circ s}\\
\]
Well:
\[
\frac{\vartheta(iw,z) + \vartheta(-iw,z)}{2} = g(w,z)
\]
Where:
\[
\sum_{k=0}^\infty f_-^{\circ 2(k+1)} \frac{(-1)^k w^{2k}}{2k!} = g(w,z)\\
\]
The function \(\frac{d^s}{dw^s} g(w,z) = (f_- \circ f_-)^{\circ s}(z)\frac{\Gamma(\frac{s}{2})\Gamma(1-\frac{s}{2})}{\Gamma(1-s)}\). Using old knowledge:
\[
\frac{d^s}{dw^s}\varphi(\lambda w) = \lambda^s \frac{d^s}{dw^s} \varphi(w)\\
\]
So we can pull out:
\[
\frac{i^s \frac{d^s}{dw^s}\vartheta(w,z) + (-i)^s \frac{d^s}{dw^s} \vartheta(w,z)}{2} = (f_- \circ f_-)^{\circ s}(z)\frac{\Gamma(\frac{s}{2})\Gamma(1-\frac{s}{2})}{\Gamma(1-s)}
\]
This means something I've worked towards for a long time. You do see this, even with the geometric case. But to pull this out in a boundary case has always escaped me. I hope I'm not fucking anything up too egregiously. I love me some hypergeometric reasoninig.
The value:
\[
f_-^{\circ s}(z) = \frac{d^s}{dw^s}\vartheta(w,z)\\
\]
Is given by the formula:
\[
f_-^{\circ 2s}(z)\cos\left(\pi s\right) = \left(f_-\circ f_-\right)^{\circ s}(z)\frac{\Gamma(s)\Gamma(1-s)}{\Gamma(1-2s)}\\
\]
(I've gotten this formula for the past year and I always thought it was too good to be true. So It might be too good to be true, but there's something like this at least appearing locally)
I hate to be the take it with a grain of salt kind of guy; but take it with a grain of salt. I might have screwed up some things.
BUT FUCKING SERIOUSLY! this my Euler equation!
\[
f_-^{\circ 2s}(z)\cos\left(\pi s\right) = \left(f_-\circ f_-\right)^{\circ s}(z)\frac{\Gamma(s)\Gamma(1-s)}{\Gamma(1-2s)}\\
\]
RE: Understanding \(f(z,\theta) = e^{e^{i\theta}z} - 1\) - MphLee - 12/28/2022
Excuse me, I hate to be that guy who ask without doing his homework... I don't get you chain of deduction, but I see the final formula and you seems to suggest that it is beautiful and also it is some analogue of Euler equation...
Can you rewrite it in a way, maybe giving a special case, that a noob like me can recognize as analogue of \(e^{ix}=\cos x +i\sin x\) or of \(e^{i\pi}=-1\)?
Also the importance of Euler identity is, secretly, that it defines a surjective group homomorphisms turning the real line into the circle... rolling it up in the unitary complex numbers....
RE: Understanding \(f(z,\theta) = e^{e^{i\theta}z} - 1\) - JmsNxn - 12/30/2022
Oh, absolutely. I can do that. Let's take \(f_\pi(z) = e^{-z}-1\).
Take the function:
\[
\vartheta_\pi(x,z) = \sum_{n=0}^\infty f_\pi^{\circ n}(z) \frac{(-x)^n}{n!}\\
\]
When I write:
\[
\vartheta_\pi(ix,z) + \vartheta_\pi(-ix,z) = 2 Q(x,z)\\
\]
This is what I mean...
\[
\vartheta_\pi(ix,z) = \sum_{n=0}^\infty f_\pi^{\circ 2n}(z) \frac{(-1)^n x^{2n}}{2n!} - i \sum_{n=0}^\infty f_\pi^{\circ 2n+1}(z) \frac{(-1)^n x^{2n+1}}{2n+1!}\\
\]
Which is just done by collecting even and odd powers...
We only want the first sum, and we want to ignore the second part. This is because the first part, looks exactly like \(\cos(x)\), but the coefficients are \(f_\pi^{\circ 2n}(z)\).
Well...
\[
\vartheta_\pi(-ix,z) = \sum_{n=0}^\infty f_\pi^{\circ 2n}(z) \frac{(-1)^n x^{2n}}{2n!} + i \sum_{n=0}^\infty f_\pi^{\circ 2n+1}(z) \frac{(-1)^n x^{2n+1}}{2n+1!}\\
\]
So that:
\[
\vartheta_\pi(ix,z) + \vartheta_\pi(-ix,z) = 2 \sum_{n=0}^\infty f_\pi^{\circ 2n}(z) \frac{(-1)^n x^{2n}}{2n!}
\]
But the equation on the right is differintegrable; because \((f_\pi \circ f_\pi)^{\circ s}\) is Mellin transformable, and Attached with the Gamma expression I wrote above, it still is. So this means the left hand side is differintegrable.
So this proves that \(f_\pi^{\circ s}(z)\) does exist, but it isn't differintegrated in a conventional manner. We obviously need to rigorify this shit some more, but the general motions look like this.
To relate this to Euler's formula... by collecting even and odd powers again...
\[
e^{ix} = \sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{2n!} + i \sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{2n+1!}\\
\]
Which famously reduces into:
\[
e^{ix} = \cos(x) + i \sin(x)\\
\]
Imagine we are doing the same thing, but we are doing something slightly different.
Let:
\[
Hu = u((f_\pi \circ f_\pi)^{\circ 1/2}) = u(h(z))\\
\]
Which is a linear operator. It takes functions holomorphic \(u(z)\) to the function \(Hu = u(h(z))= u((f_\pi \circ f_\pi)^{\circ 1/2}(z))\). (Again, you have to note that \(h = (f_\pi \circ f_\pi)^{\circ 1/2} \neq f_\pi\), the left element has \(1\) as a multiplier at \(0\), and the right has \(-1\) as a multiplier at \(0\))... This is the exact same equation as \(-1 = \sqrt{(-1)(-1)} = 1\). We have to pay attention how we take the root; and be very careful.
To justify it's a linear operator. Let \(\alpha, \beta \in \mathbb{C}\). Let \(u,v : \mathbb{C} \to \mathbb{C}\). Then:
\[
H\left(\alpha u(z) + \beta v(z)\right) = \alpha u(h(z)) + \beta v(h(z)) = \alpha Hu + \beta Hv\\
\]
Then:
\[
\cos(Hx)z = \sum_{n=0}^\infty H^{2n}z \frac{(-1)^n x^{2n}}{2n!} = \sum_{n=0}^\infty f_\pi^{\circ 2n}(z) \frac{(-1)^n x^{2n}}{2n!}\\
\]
This function is differintegrable in \(x\), because:
\[
e^{Hx}z = \sum_{n=0}^\infty H^n \frac{x^n}{n!}\\
\]
is differintegrable; and by a change of variables I described with the \(\Gamma\) function, we know that both:
\[
\cos(Hx)z + i\sin(Hx)z = e^{iHx}z\\
\]
are differintegrable. So now, we're only going to focus on \(\cos\) (we could focus on \(\sin\) just as well, but \(\cos\) is easier). If I make a different operator \(P\), such that:
\[
Pu = u(f_\pi(z))\\
\]
The function:
\[
\cos(Px)z = \cos(Hx)z\\
\]
So that, even though \(e^{-Hx} \neq e^{-Px} = \vartheta_\pi(x)\)--and \(e^{Px}\) is only differintegrable in a distributional sense (where it is the solution of an algebraic equation extending the integral transform), we can successfully take:
\[
\frac{d^{s}}{dx^{s}} e^{Px}\\
\]
Where, all we have to do is take:
\[
\frac{d^s}{dx^s} \frac{e^{iPx} + e^{-iPx}}{2} = \frac{d^s}{dx^s} \cos(Px) = \frac{d^s}{dx^s} \cos(Hx)\\
\]
Out pops a bunch of Gamma nonsense that looks like something from Riemann's journal (Again, I'm a big analytic number theory guy, lol). Exactly what pops out, is what you saw above (I may have fudged some subtleties, this stuff still gives me a headache, but it's something like this). But in the distributional sense:
\[
\frac{d^s}{dx^s} \frac{e^{iPx} + e^{-iPx}}{2} = P^s \frac{i^s + i^{-s}}{2} = P^s \frac{e^{i\pi s/2} + e^{-i \pi s/2}}{2} = P^s \cos(\pi s /2) = H^s \frac{\Gamma(s/2)\Gamma(1-s/2)}{\Gamma(1-s)}\\
\]
I expect \(P^s = \frac{d^s}{dx^s} e^{Px} = \frac{d^s}{dx^s} \vartheta_\pi(x)\) to actually be pretty poorly behaved. If it was super well behaved, then \(\vartheta_\pi\) wouldn't need distributional analysis. So \(P^s\) may have singularities; or malaligned behaviour. I'm going to have to look at it much deeper. I'm just giving a sketch for now.
So essentially, the point is, under the inverse Mellin transform as a distributional transform (distributional Fourier transform), we have the equality:
\[
P^s \cos(\pi s /2) = H^s \frac{\Gamma(s/2)\Gamma(1-s/2)}{\Gamma(1-s)}\\
\]
So even though the Mellin transform of \(\vartheta_\pi\) doesn't converge; some extension of the Mellin transform of \(\vartheta_\pi\) does (advanced functional analysis shit, I'm going to have to put on my thinking cap). And the converged version is the above formula (forgoing I didn't forget a negative sign here or there in my analysis
)Regards, James
EDIT:
Please remember that \(P\) and \(H\) are linear operators. But additionally they are actionable under a Fourier like transform. Additionally \(P^2 = H^2\). This means, we can construct a Hilbert space (In our case it's just looks like a Hardy space). We can look at these things like Gottfried (Heisenberg), and \(P,H\) are just infinite square matrices acting on an infinite vector space. We can look at this like waves, and that we are altering frequencies of the waves through \(P,H\)... Schrodinger shit. Or we can collectively say all of us are Von Neumann, and we're going to hands down prove both of these interpretations are the same
.Let's do some Von Neumann. I believe I can generate an asymptotic series at \(z\approx 0\) for \(f_\pi^{\circ 1/2}(z) =g(z)\) which starts with the leading term \(g(z) = iz + O(z^2)\). Just as \(\sqrt{-1} = i\)
The choice is ours.
RE: Understanding \(f(z,\theta) = e^{e^{i\theta}z} - 1\) - JmsNxn - 01/01/2023
Alright; so it appears I made a fumble. The Gamma functions should almost surely cancel out in some manner; so they appear in the process, but probably won't make an appearance in the final formula. To justify this, let's look at:
\[
f_\theta(z) = e^{e^{i\theta}z} -1\\
\]
But we're going to restrict \(\Im\theta > 0\); and then limit \(\Im\theta \to 0\), to hit the neutral case.
When \(\Im\theta > 0\), the function \(f_\theta(z) = e^{i\theta} z + O(z^2)\) and \(|e^{i\theta}| < 1\); so this is a geometrically attracting fixed point at \(z=0\). This means we can write:
\[
\Psi_\theta(f_\theta(z)) = e^{i\theta} \Psi_\theta(z)\\
\]
For a unique Schroder function \(\Psi_\theta\). Where then the regular iteration is:
\[
f_\theta^{\circ s}(z) = \Psi_\theta^{-1} \left( e^{i\theta s} \Psi_\theta(z)\right)\\
\]
Which is holomorphic for \(|z| < \delta\) and \(|e^{i\theta s}| < 1\)--which is a half plane in \(s\) if we fix \(\theta\).
We are going to focus on \(f_\pi(z) = e^{-z} -1\), but we're going to instead write \(g_n(z) = e^{-z + z/n} -1\). The function \(g_n(z) = (\frac{1}{n} -1) z + O(z^2)\), so that it has an attracting fixed point with multiplier \(\lambda_n = 1/n - 1\). Where the limit \(g_n \to f_\pi\) as \(n\to \infty\). This can also be written as \(f_{\theta_n}(z)\) when \(e^{i\theta_n} = 1/n - 1\)--which is \(\theta_n = -i\log(1/n - 1)\).
The function \(g_n(g_n(z)) = h(z)\), and \(h(z)\) looks like:
\[
h(z) = (1/n-1)^2 z + O(z^2)\\
\]
And thereby; if we write:
\[
\vartheta[h](x,z) = \sum_{k=0}^\infty h^{\circ k}(z) \frac{(-x)^k}{k!}\\
\]
Then:
\[
\frac{d^{s}}{dx^s}\Big{|}_{x=0} \vartheta[h](x,z) = h^{\circ s}(z)\\
\]
But this is not an iteration of \(g_n\)! Capitulated by the statement \(\sqrt{h} \neq g_n\)....
So how do we make the square root of \(h\) equal \(g_n\)? How do we take "the negative square root".
This is an old idea of mine, I had never fully developed. But we return now to the Schroder expansion:
\[
h(z) = \Psi_{\theta_n}^{-1}\left( (1/n-1)^2 \Psi_{\theta_n}(z)\right)\\
\]
Whereupon, the formula we have given for \(h^{\circ s}(z)\) is aptly written:
\[
h^{\circ s}(z) = \Psi_{\theta_n}^{-1}\left( |1/n-1|^{2s} \Psi_{\theta_n}(z)\right)\\
\]
If we wanted \(h^{\circ 1/2} (z) = g_n(z)\), we would need to stick a negative somewhere in there. We do this by massaging our exponents.
\[
(1/n-1)^{2s} = e^{2\pi i s} |1/n-1|^{2s} = |1/n-1|^{2s +\frac{2 \pi i}{\log|1/n-1|}}\\
\]
Whereby; we can now deduce that:
\[
h^{\circ \frac{1}{2} + \frac{\pi i}{\log|1/n-1|}}(z) = g_n(z)\\
\]
Where now we can iterate \(g_n(z)\) (not quite under the mellin transform, but close) by writing:
\[
g_n^{\circ s}(z) = h^{\circ \frac{s}{2} + \frac{\pi i s}{\log|1/n-1|}}(z)\\
\]
(THIS CHANGE OF VARIABLES I FORGOT WILL CAUSE THE GAMMA FUNCTIONS TO JUST DISAPPEAR).
I'm going to walk through a simple case here, and explain how we are doing the same thing. This result dates very far back; but probably was never spoke of in this context.
When we take the function:
\[
e^{x}\\
\]
Then the Exponential Differintegral, is precisely itself:
\[
\frac{d^s}{dx^s} e^x = e^x\\
\]
When we change to \(e^{-x}\); then something a little tricky happens...
First of all; The exponential differintegral is defined as follows:
\[
\frac{d^{-s}}{dx^{-s}} f(x) = \frac{e^{-i\tau s}}{\Gamma(s)} \int_0^\infty f(x - e^{i\tau}y)y^{s-1}\,dy\\
\]
Where if a function is differintegrable, there is some open sector \(-\pi \le a < \tau < b \le \pi\), where this thing converges (And it's invariant under our choice of \(\tau\), constant in this sector--thanks to Cauchy). Well, for \(f(x) = e^{-x}\), this happens at \( \tau = \pi\). (It happens as we move in the opposite direction). So that:
\[
\frac{d^{-s}}{dx^{-s}} e^{-x} = \frac{e^{-i\pi s}}{\Gamma(s)} \int_0^\infty e^{-(x-e^{i\pi}y)}y^{s-1}\,dy = e^{-\pi i s} e^{-x}\\
\]
Now we do all of this solely with the principle branch; choosing different solutions to the differintegral outside of the principle branch equates to multiplying by \(e^{2 \pi i k s}\) for some \(k \in \mathbb{Z}\), so bear with me...
This looks like a non result; but it perfectly aligns the equation:
\[
\frac{d^s}{dx^s} e^{\mu x} = \mu^s e^{\mu x}\\
\]
And does so in the complex plane; and does so while accounting for different paths of convergence for each \(\mu\). Where the Exponential Differintegral is not just a mellin transform, it's precisely a generalized mellin transform that behaves as an Extension of a Linear Operator in a Hilbert space...We are extending a Hilbert Space a la Von Neumann. This is just an example of this.
So returning to the problem at hand, we can write:
\[
g_n^{\circ s}(z) = \frac{1}{\Gamma(s)} \int_0^\infty \vartheta[h](x,z)x^{-\frac{s}{2} -\frac{\pi i s}{\log|1/n-1|}-1}\,dx\\
\]
This will not converge on a maximal domain; but if we instead write:
\[
g_n^{\circ s}(z) = \frac{d^{\frac{s}{2} +\frac{\pi i s}{\log|1/n-1|}}}{dx^{\frac{s}{2} +\frac{\pi i s}{\log|1/n-1|}}}\Big{|}_{x=0} \vartheta[h](x,z)
\]
We can achieve the exact same maximal domain as:
\[
g_n^{\circ s}(z) = \Psi_{\theta_n}^{-1} \left( (1/n - 1)^{s} \Psi_{\theta_n}(z)\right)\\
\]
This will produce a Mittag Leffler expansion I'm too lazy to write out; but it'll look a lot like what I wrote above. Essentially, the \(\frac{d^s}{dx^s}\) is holomorphic for \(\Re(s) > 0\), but with this modification we have to map that domain to a new one.
This kind of ties the knot to what I was saying before. Let's try to find the iterate of \(g_n\) using the vartheta function of \(g_n\).
\[
\vartheta[g_n](x,z) = \sum_{k=0}^\infty g_n^{\circ k}(z) \frac{(-x)^k}{k!}\\
\]
Then! Since \(g_n^{\circ 2k}(z) = h^{\circ k}(z)\); we have that:
\[
\frac{\vartheta[g_n](ix,z) + \vartheta[g_n](-ix,z)}{2} = \frac{\vartheta[\sqrt{h}](ix,z) + \vartheta[\sqrt{h}](-ix,z)}{2}
\]
What we should end up with is going to look a lot wackier though; because, in basic principle:
\[
\frac{d^s}{dx^s} \vartheta[g_n](x,z) = g^{\circ s}(z) = h^{\circ \frac{s}{2} + \frac{\pi i s}{\log|1/n-1|}}(z)
\]
This may appear to be a contradiction, but:
\[
h^{\circ s + \frac{2\pi i}{\log|1/n-1|}}(z) = h^{\circ s}\\
\]
So we are utilizing a duplication formula, and the halving of a period, which when duplicated becomes the same function. And we can argue canonically that \(g_n^{\circ 2k} = h^{\circ k}\), with absolutely zero repercussions on our iterated format.
So This actually causes the Gamma function stuff to disappear
; as it should appear on both sides of the equation, and what we our left with is some kind of translation in the original argument.When we let \(\Im\theta \to 0\), things are definitely going to get complicated. As \(\theta_n \to \pi\), something more clever is needed to do these arguments. But something like it should appear.
To explain where I went wrong is a little tricky, but I'll go as follows:
\[
\frac{1}{2\pi i} \int_{1/2 - i \infty}^{1/2 + i \infty} h^{\circ s}(z) \frac{\Gamma(s)\Gamma(1-s)}{\Gamma(1-2s)}x^{-s}\,ds = \sum_{k=0}^\infty h^{\circ k}(z) \frac{(-x)^{k}}{2k!}\\
\]
But,
\[
\frac{\vartheta[g_n](i\sqrt{x},z) + \vartheta[g_n](-i\sqrt{x},z)}{2} = \sum_{k=0}^\infty h^{\circ k}(z) \frac{(-x)^{k}}{2k!}\\
\]
From this point, we should just end up with the formula:
\[
g_n^{\circ s} = h^{\circ \frac{s}{2} + \frac{\pi i s}{\log|1/n-1|}}\\
\]
........ The gamma's should cancel out....
Regards, James
RE: Understanding \(f(z,\theta) = e^{e^{i\theta}z} - 1\) - JmsNxn - 01/01/2023
I'm going to continue this train of thought now, while I have it.
This essentially reduces "under the mellin transform" transformations, into the standard "regular iteration" but only for specific values. Namely, when:
\[
\Im \theta >0\,\,\Re \theta = k \pi\,\,\text{for}\,k \in \mathbb{Z}\\
\]
Because, \(f_\theta\) in this instance, satisfies \(0 < e^{i2\theta} < 1\)--is real positive.
But there is nothing stopping us from extending this for all \(\Im \theta > 0\). We just have to think a bit more clever. Let's take:
\[
\theta \in 2\pi \mathbb{Q}(i)\\
\]
Where, still, \(\Im(\theta) > 0\). So that \(\theta = 2 \pi a + 2 \pi b i\), where \(a,b \in \mathbb{Q}\) and \(b > 0\).
And let's assume that \( q \theta = \in 2\pi \mathbb{Z}(i)\), and \(q \in \mathbb{N}\) is the minimal element to do so. The set \(2 \pi \mathbb{Q}(i)\) is dense in \(\Im(\theta) > 0\). This is nothing new.
The function \(f_\theta^{\circ q}(z) = h(z)\). Then... \(h'(0) = e^{i\theta q}\), satisfies, \(0 < e^{i \theta q} < 1\) is real positive.
This means that \(h^{\circ s}\) has the same period as \(e^{i\theta q s}\), which is precisely \(\frac{2 \pi}{\theta q}\). We are taking a \(q\)'th of this, which says that:
\[
f_\theta^{\circ s}(z) = h^{\circ \frac{s}{q} + \frac{2 \pi s}{q^2 \theta}}(z)\\
\]
BUT!!!!!!! Here is the huge fucking but. When we take a square root, like I did in my last post; we only have \(1\) or \(-1\). In this case, we have precisely \(q\) roots. Since we have chosen that \(q\) is the minimal value for \(q \theta \in 2 \pi \mathbb{Z}(i)\)--there are \(q\) other values which satisfy this equation. (The above equation is fine, but it can be tricky to answer...) So the correct answer, and the more blanket version is...
There exists some \(0 \le t < q\) such that:
\[
f_\theta^{\circ s}(z) = h^{\circ \frac{s}{q} + \frac{2 \pi s t}{q^2 \theta}}(z)\\
\]
Now each one is not a solution, but sum are (Fucking Gaussian sum shit); but we don't care. We've found the Regular iteration solution.
So:
\[
f_\theta^{\circ s}(z) = h^{\circ \frac{s}{q} + \frac{2 \pi s}{q^2 \theta}}(z)\\
\]
Where: \(h(z) = f_\theta^{\circ q}(z)\)
Where \(q \in \mathbb{N}\) is the minimal integer, so that \(q\theta \in 2\pi \mathbb{Z}(i)\)...
Again........ the Gamma's cancel out, but in the Mellin transform we use Gauss's Gamma multiplication formula....
We can also see a cool looking Gauss like sum kinda appearing.....
We're going to have a lot of trouble taking \(\Im\theta \to 0\). But honestly, not that much trouble. So long as we don't send \(\theta \to 2 \pi k\) we should be fine. So the singularity action is precisely at \(f_0(z) = e^z -1\), which we know how to iterate. At \(f_\pi(z) = e^{-z} -1\) we're not gonna see such a singularity. It should behave better-- and near zero we can accurately say that:
\[
f_\theta^{\circ s}(z) = e^{i\theta s}z + O(z^2)\\
\]
As we limit \(\Im \theta \to 0\), this becomes an asymptotic series; so it is not holomorphic. But it still looks like this expression.
THE BIG WHODUNNIT ABOUT THIS IS THAT WE CAN WRITE THIS UNDER A MELLIN TRANSFORM.
\[
\int_{1/2 - i\infty}^{1/2 + i\infty} h^{\circ s/q}(z) \frac{\Gamma(s/q)\Gamma(1-s/q)}{\Gamma(1-s)}x^{-s}\,dx = \int_{1/2 - i \infty}^{1/2+i\infty} f_\theta^{\circ s}(z)\frac{\Gamma(s/q)\Gamma(1-s/q)}{\Gamma(1-s)}x^{-s}\,dx\\
\]
Which is basically the statement that:
\[
\sum_{k=0}^\infty h^{\circ k}(z) \frac{(-x)^k}{(qk)!} = \sum_{k=0}^\infty f_\theta^{\circ qk}(z) \frac{(-x)^k}{(qk)!}\\
\]
But now we've written it with the Fourier transform, but not just the Fourier transform; a Distributional understanding of the Fourier Transform...
But this only happens in a distributional sense. The right hand integral does not converge \(f_\theta^{\circ s}\) does not decay properly. But there's only a single other function which satisfies this in a distributional sense (once we've extended the Fourier transform linearly).... \(f_\theta^{\circ s} = h^{\circ \frac{s}{q} + \frac{2 \pi t}{q^2 \theta}}\) for \(t\) coprime to \(q\)--but the regular iteration is given as \(t =1\). The other iterations will fail a unification, monodromy theorem, but will act as mock solutions. As they will not give \(f'_\theta(0) = e^{i\theta}\).
So....... the Gamma function stuff I wrote above, it actually cancels out.
I apologize. I saw some Gammas and got excited and ahead of myself lmao.
RE: Understanding \(f(z,\theta) = e^{e^{i\theta}z} - 1\) - JmsNxn - 01/03/2023
So, let's write:
\[
f_\theta^{\circ s}(z) = h^{\circ \frac{s}{q} + \frac{2 \pi s}{q^2\theta}}\\
\]
This can be writ, as:
\[
f_\theta^{\circ -s}(z) = \frac{1}{\Gamma(s)} \int_0^\infty \vartheta[h](x,z) x^{\frac{s}{q} + \frac{2 \pi s}{q^2\theta} -1}\,dx\\
\]
Let \(u = x^{\frac{1}{q} + \frac{2 \pi}{q^2\theta}}\) where by \(du = \left(\frac{1}{q} + \frac{2 \pi}{q^2\theta}\right)x^{\frac{1}{q} + \frac{2 \pi}{q^2\theta}-1}\,dx\). Whereby we now have the equation:
\[
f_\theta^{\circ -s}(z) = \frac{1}{(\frac{1}{q} + \frac{2 \pi}{q^2\theta})\Gamma(s)}\int_0^\infty \vartheta[h](u^{q + \frac{q^2}{2\pi\theta}},z) u^{s-1}\,du\\
\]
Thereby, we can write:
\[
H(x,z) = \frac{1}{\frac{1}{q} + \frac{2 \pi}{q^2\theta}}\sum_{k=0}^\infty f_\theta^{\circ kq}(z) \frac{(-1)^kx^{(q + \frac{q^2}{2\pi\theta})k}}{k!}\\
\]
And we should be able to derive that:
\[
\frac{d^s}{dx^s}\Big{|}_{x=0} H(x,z) = f_\theta^{\circ s}(z)\\
\]
.......... Provided I didn't screw up my elementary calculus.... We may have to take this integral along the arc \([0, \lim_{x\to\infty}x^{1/\theta}]\), rather than just along the real line....
God, I haven't had to keep track of this many variable substitutions since like 2nd or 3rd year undergrad. Getting frustrated....
Either way, it's something like this, lol.Oddly enough; the value:
\[
\frac{1}{\frac{1}{q} + \frac{2 \pi}{q^2\theta}} \approx \frac{1}{m}\\
\]
where \(m\to\infty\) as \(q \to \infty\). But additionally:
\[
f_\theta^{\circ kq} \frac{(-1)^kx^{(q + \frac{q^2}{2\pi\theta})k}}{k!} \approx f_\theta^{kq} \frac{(-1)^k x^{kq}}{k!}\\
\]
So that:
\[
\frac{1}{m} \sum_{k=0}^\infty f_\theta^{kq/m} \frac{(-1)^k x^{kq/m}}{k!} \approx \int_0^\infty f^{\circ kq}(z) \frac{e^{\pi i k}x^{kq}}{k!}\,dk\\
\]
I smell a fourier transform somewhere in here. Too tired tn, but I think we can solve for when \(q \to \infty\) using integral calculus. Which would allow us to talk about the neutral case... when \(\Im\theta \to 0\) and \(q = \infty\)...
Recalling that all the work of this post, only works for the set \(\theta \in 2 \pi \mathbb{Q}(i) \cap \Im(\theta) > 0\), and nowhere else. We have to take limits to get it to work for \(\Im(\theta) \ge 0\) for arbitrary functions.
RE: Understanding \(f(z,\theta) = e^{e^{i\theta}z} - 1\) - JmsNxn - 01/03/2023
Let's write:
\[
f_\theta^{\circ -s}(z) = \frac{1}{\Gamma(s)} \int_\gamma H(x,z)x^{s-1}\,dx\\
\]
When:
\[
f_\theta(z) = e^{e^{i\theta}z}-1\\
\]
We assign the following linear relationships:
\[
\begin{align}
e^{i\theta q} &\in (0,1)\\
\theta q &\in 2 \pi \mathbb{Z}(i)\\
q \,&\text{is the minimal value}\, q \in \mathbb{N}\,\text{of all solutions}\\
\Im(\theta) &> 0\\
\theta &\in 2\pi \mathbb{Q}(i)\\
\end{align}
\]
Then...
\[
H(x,z) = \frac{1}{\frac{1}{q} + \frac{2\pi}{q^2\theta}} \sum_{k=0}^\infty f_\theta^{\circ kq}(z) \frac{(-1)^k x^{(q+\frac{q^2\theta}{2\pi})k}}{k!}\\
\]
Here enters the integral...
Let \(h = \frac{1}{q} + \frac{2\pi}{q^2\theta}\).
As \(h \to 0\) we have \(q \to \infty\). This happens exactly as \(\theta \to \mathbb{C}(i)\), rather than \(\theta \in 2\pi \mathbb{Q}(i)\). As we hit limit points, the value \(q \to \infty\)...
This means the expression:
\[
x^{(q+\frac{q^2\theta}{2\pi})k} = x^{k/h}\\
\]
This is close enough to pull a proper inspection--and only be a linear factor off...
From here we have:
\[
f_\theta^{\circ kq} = f_\theta^{\circ \frac{k}{h}}\\
\]
This is because we only have added a second order error to \(h\), and the second order error will make no demonstration here.
Let's run the limit formula using these constraints...
\[
H(x,z) = \lim_{h\to 0} \frac{1}{h} \sum_{k=0}^\infty f_\theta^{\circ \frac{k}{h}} \frac{(-1)^k x^{\frac{k}{h}}}{k!}\\
\]
This is close to the integral we actually want. It will not be the integral. But it's reallllllyyyyyyyy close.
I CAN FUCKING SMELL FOURIER BUT I KNOW I'M MISSING SOMETHING
Okay, so I know that :
\[
H(x,z) = \int_0^\infty f_\theta^{\circ t} x^t\,d\mu\\
\]
Where \(\mu(t)\) is a measure on \([0,\infty]\). It's something like this this.... Fuck boys! I may have dug a hole too deep!!!!